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Question: Bromine monochloride, \( BrCl \) decomposes into bromine and chlorine and reaches the equilibrium. ...

Bromine monochloride, BrClBrCl decomposes into bromine and chlorine and reaches the equilibrium.
2BrCl(g)⇄Br2(g)+Cl2(g)2BrC{l_{(g)}} \rightleftarrows B{r_{2(g)}} + C{l_{2(g)}} for which Kc=32{K_c} = 32 at 500K500K .
If initially pure BrClBrCl is present at a concentration of 3.30×10−3mollitre−13.30 \times {10^{ - 3}}mollitr{e^{ - 1}} ; what is its molar
concentration in the mixture at equilibrium?

Explanation

Solution

Hint : To calculate an equilibrium concentration from an equilibrium constant, an understanding of the concept of equilibrium and how to write an equilibrium constant is required. Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.

Complete Step By Step Answer:
For knowing the value of molar concentration BrClBrCl we need to first calculate the degree at which it gets decomposed, that is the degree of dissociation of BrClBrCl into its constituent ions.
let the degree of dissociation be α\alpha .
We observe the decomposition reaction at initial concentration and at equilibrium.
2BrCl(g)⇄Br2(g)+Cl2(g)2BrC{l_{(g)}} \rightleftarrows B{r_{2(g)}} + C{l_{2(g)}}
3.30×10−33.30 \times {10^{ - 3}} 0 0
3.3×10−3−2α3.3 \times {10^{ - 3}} - 2\alpha α\alpha α\alpha
At the initial stage only BrClBrCl is present and its concentration is 3.30×10−3mol litre−13.30 \times {10^{ - 3}}mol{\text{ }}litr{e^{ - 1}} , but the concentration of ions will be zero.
At the equilibrium dissociation occurs and equal molecules of ions are formed that is α\alpha and 2 α\alpha amount is reduced from the concentration of BrClBrCl .
Now Kc{K_c} of the reaction at equilibrium is given by:
Kc=[Br2][Cl2][BrCl2]2{K_c} = \dfrac{{\left[ {B{r_2}} \right]\left[ {C{l_2}} \right]}}{{{{\left[ {BrC{l_2}} \right]}^2}}}
Substituting the respective concentration in the above equations:
Kc=α×α(3.3×10−3−2α)2{K_c} = \dfrac{{\alpha \times \alpha }}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}}
Kc=α2(3.3×10−3−2α)2{K_c} = \dfrac{{{\alpha ^2}}}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}}
Kc=32{K_c} = 32 (given)
α2(3.3×10−3−2α)2=32\dfrac{{{\alpha ^2}}}{{{{(3.3 \times {{10}^{ - 3}} - 2\alpha )}^2}}} = 32
Taking square root both sides we get:
α3.3×10−3−2α=42\dfrac{\alpha }{{3.3 \times {{10}^{ - 3}} - 2\alpha }} = 4\sqrt 2
α=18.67×10−3−82α\alpha = 18.67 \times {10^{ - 3}} - 8\sqrt 2 \alpha
α+82α=18.67×10−3\alpha + 8\sqrt 2 \alpha = 18.67 \times {10^{ - 3}}
(1+82)α=18.67×10−3\left( {1 + 8\sqrt 2 } \right)\alpha = 18.67 \times {10^{ - 3}}
α=1.5162×10−3\alpha = 1.5162 \times {10^{ - 3}}
Hence now we have the value of degree of dissociation.
So, Molar Concentration of BrClBrCl at equation:
That is given by
3.3×10−3−2α3.3 \times {10^{ - 3}} - 2\alpha
3.3×10−3−2×1.5162×10−33.3 \times {10^{ - 3}} - 2 \times 1.5162 \times {10^{ - 3}} = 0.267×10−3M0.267 \times {10^{ - 3}}M
Hence the required answer is: 0.267×10−3M0.267 \times {10^{ - 3}}M .

Note :
Molar concentration (also called molarity, amount concentration or substance concentration) is a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.